What is the value of $a^2+b^2+c^2+d^2$ and $a^3+b^3+c^3+d^3$ for $a,b,c,d$ roots of $f(x)=(x-1)^2(x-2)^2+1$

Question

My idea was $f(x)=0 \iff (x-1)^2(x-2)^2=-1$, so $(x-1)(x-2)=i$ but this leads to solutions of the form $x=\frac{1}{2}(3\pm \sqrt{1\pm 4i})$. Now I could tediously calculate the sum of squares by hand, however the sum of cubes is out of reach with this method. The problem was part of an old exam and should be solvable without too much calculations in a couple of minutes. What trick am I missing?

Answer 1

Knowing the expanded form :

$$x^4 - 6x^3 + 13x^2 - 12x + 5$$

of the polynomial, with roots denoted $r_k$, use Newton identities that you will find here :(https://en.wikipedia.org/wiki/Newton%27s_identities) : for the sum of squares of roots :

$$p_2=e_1^2-2e_2$$

and sum of cubes of roots :

$$p_3=e_1^3-3e_1e_2+3e_3$$

where the $e_k$ are the so-called "elementary polynomials" :

$$\begin{cases} e_1&=&r_1+r_2+r_3+r_4&=&6\\ e_2&=&r_1r_2+r_1r_3+...+r_3r_4&=&13\\ e_3&=&r_1r_2r_3+r_1r_2r_4+... r_2r_3r_4&=&12 \end{cases}$$

where the last equalities are due to Viète formulas.

Answer 2

HINT.-We have $$(x^2-3x+2)^2=-1\Rightarrow x^2-3x+(2\mp i)=0$$ from which $$2a=3+\sqrt{1+4i}\\2b=3-\sqrt{1+4i}\\2c=3+\sqrt{1-4i}\\2d=3-\sqrt{1-4i}$$ Therefore $$4(a^2+b^2+c^2+d^2)=(9+(1+4i)+6\sqrt{1+4i})+(9+(1+4i)-6\sqrt{1+4i})+(9+(1-4i)+6\sqrt{1+4i})+(9+(1-4i)-6\sqrt{1-4i})$$ this sum gives $$4(a^2+b^2+c^2+d^2)=40$$ Similarly we have $$(3\pm\sqrt{1\pm4i})^3=36\pm27\sqrt{1\pm4i}\pm(1\pm4i)\pm36i$$ then $$8(a^3+b^3+c^3+d^3)=4\times36$$ Finally we get $$a^2+b^2+c^2+d^2=10\\a^3+b^3+c^3+d^3=18$$

Answer 3

The roots of $(x-1)^2(x-2)^2=-1$ are $x_k$, $k=1,4$, where $x_{1,2}$ are the roots of $(x-1)(x-2) = i$, and $x_{3,4}$ are the roots of $(x-1)(x-2) = -i$. We can take $x_3= \bar x_1$, $x_4=\bar x_2$. Now $(x-1)(x-2)-i= x^2-3 x +2-i$, so $$x_1^2+x_2^2=(x_1+x_2)^2-2 x_1 x_2 =3^2-2(2-i) = 5+2i$$ so $x_3^2+x_4^2=5-2i$, and $\sum x_k^2=10$.

Also, $$x_1^3+x_2^3=(x_1+x_2)^3-3 (x_1+x_2)x_1 x_2=3^3-3\cdot 3(2-i)=9+ \ldots$$ so $x_3^3+x_4^3=9-\ldots$, and $\sum x_k^3= 18$.